2.3 Linear Algebra 75ui=∏ 3
∏k=^1 (ai+xk)
k =i(ai−ak),i= 1 , 2 , 3.It is not hard to check that these satisfy the system of equations. Observe that
∂ui
∂xj=
∏
∏k =j(ai+xk)
j =i(ai−aj), and so∂ui
∂xj=
1
ai+xj
ui,i,j= 1 , 2 , 3.The determinant in question looks again difficult to compute. Some tricks simplify the
task. An observation is that the sum of the columns is 1. Indeed, these sums are
∂u 1
∂xi+
∂u 2
∂xi+
∂u 3
∂xi
,i= 1 , 2 , 3 ,which we should recognize as the left-hand sides of the linear system. So the determinant
becomes much simpler if we add the first and second rows to the last. Another observation
is that the determinant is a 3-variable polynomial inx 1 ,x 2 ,x 3. Its total degree is 3, and it
becomes zero ifxi=xjfor somei =j. Consequently, the determinant is a number not
depending onx 1 ,x 2 ,x 3 times(x 1 −x 2 )(x 2 −x 3 )(x 3 −x 1 ). This number can be determined
by looking just at the coefficient ofx^22 x 3. And an easy computation shows that this is
equal to(a 1 −a 2 )(a 2 −^1 a 3 )(a 3 −a 1 ).
From the very many practical applications of the theory of systems of linear equations,
let us mention the Global Positioning System (GPS). The principle behind the GPS is
the measurement of the distances between the receiver and 24 satellites (in practice some
of these satellites might have to be ignored in order to avoid errors due to atmospheric
phenomena). This yields 24 quadratic equationsd(P,Si)^2 =ri^2 ,i= 1 , 2 ,...,24, in
the three spatial coordinates of the receiver. Subtracting the first of the equations from
the others cancels the quadratic terms and gives rise to an overdetermined system of 23
linear equations in three unknowns. Determining the location of the receiver is therefore
a linear algebra problem.
231.Solve the system of linear equations
x 1 +x 2 +x 3 = 0 ,
x 2 +x 3 +x 4 = 0 ,
···
x 99 +x 100 +x 1 = 0 ,
x 100 +x 1 +x 2 = 0.232.Find the solutionsx 1 ,x 2 ,x 3 ,x 4 ,x 5 to the system of equations
x 5 +x 2 =yx 1 ,x 1 +x 3 =yx 2 ,x 2 +x 4 =yx 3 ,